Question: We know that $~\frac{1}{1+x}=1-x+{{x}^{2}}-...+{{\left( -1 \right)}^{n}}{{x}^{n}}+...$ for $x\in (-1,1)$. Using this fact, find the power series for $\dfrac{1}{(1+x)^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}+...$ (Choice B) B $-x+\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}+...$ (Choice C) C $-1+2x-3{{x}^{2}}+...$ (Choice D) D $1-2x+3{{x}^{2}}+...$
First, take the derivative with respect to $~x~$ of both sides of the given equation. ${{D}_{x}}\left( \frac{1}{1+x} \right)={{D}_{x}}\left( 1-x+{{x}^{2}}-{{x}^{3}}+... \right)$ This gives the following equality. $-\frac{1}{{{\left( 1+x \right)}^{2}}}=-1+2x-3{{x}^{2}}+...$ Multiplying both sides by $~-1~$ yields the desired result. $\frac{1}{{{\left( 1+x \right)}^{2}}}=1-2x+3{{x}^{2}}-...$